Question

Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.1 kg⋅m2 and for arms and legs in is 0.90 kg⋅m2 . If she starts out spinning at 4.0 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?

Answer 1

The angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s

Step-by-step explanation:

Given;

moment of inertia of a skater with arms out,
`I_(arms \ out)` = 3.1 kg.m²

moment of inertia of a skater with arms in,
`I_(arms \ in)` = 0.9 kg.m²

inward angular speed,
`\omega _(in)` = 4 rev/s

The angular momentum of the skater when her arms are out and one leg extended is equal to her angular momentum when her arms and legs are in.

`L_(out) = L_(in)`

`I_(out) \omega_(out) = I_(in) \omega_(in)\\\\\omega_(out) = ( I_(in) \omega_(in) )/(I_(out) ) \\\\\omega_(out) = (0.9*4)/(3.1) \\\\\omega_(out) = 1.161 \ rev/s`

Therefore, the angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s